Mathematics, 27.07.2019 19:30 werdtotheblue691

The length of the base edge of a pyramid with a regular hexagon base is represented as x. the height of the pyramid is 3 times longer than the base edge. the height of the pyramid can be represented . the of an equilateral triangle with length x is units2. the area of the hexagon base times the area of the equilateral triangle. the volume of the pyramid x3 units3.1. a 3 b 3+x c 3x d 92 a altitude b apothem c area d volume 3 a 2 b 3 c 4 d 64 a 3/2 b 3 c 6 d 9/2

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Answer from: SoccerAllStar2

(a)

h=3x

(b)

$A=\frac{\sqrt{3} }{4} x^2$

(c)

$A=\frac{3\sqrt{3} }{2} x^2$

(d)

$V=\frac{3\sqrt{3} }{2} x^3$ units^3

Step-by-step explanation:

We are given a regular hexagon pyramid

Since, it is regular hexagon

so, value of edge of all sides must be same

The length of the base edge of a pyramid with a regular hexagon base is represented as x

so, edge of base =x

b=x

Let's assume each blank spaces as a , b , c, d

we will find value for each spaces

(a)

The height of the pyramid is 3 times longer than the base edge

so, height =3*edge of base

height=3x

h=3x

(b)

Since, it is in units^2

so, it is given to find area

we know that

area of equilateral triangle is

$=\frac{\sqrt{3} }{4} b^2$

h=3x

b=x

now, we can plug values

$A=\frac{\sqrt{3} }{4} x^2$

(c)

we know that

there are six such triangles in the base of hexagon

So,

Area of base of hexagon = 6* (area of triangle)

Area of base of hexagon is

$=6\times \frac{\sqrt{3} }{4} x^2$

$=\frac{3\sqrt{3} }{2} x^2$

(d)

Volume=(1/3)* (Area of hexagon)*(height of pyramid)

now, we can plug values

Volume is

$=\frac{1}{3}\times\frac{3\sqrt{3} }{2} x^2\times (3x)$

$V=\frac{3\sqrt{3} }{2} x^3$ units^3

Answer from: coxtinam16

The true statements are:

The height of the pyramid can be represented as 3x. The area of an equilateral triangle with length x is $\frac{\sqrt 3}{4}x^2$ units^2.The area of the hexagon base is six times the area of the equilateral triangle. The volume of the pyramid is $\frac{3\sqrt 3}{2}x^3$ units^3.

The length of base edge is given as:

Length=x

The height is three times the length.

So, the expression that represents the height is:

Height =3x

The area of an equilateral triangle is then calculated as:

$A_t = \frac{\sqrt 3}{4}l^2$

Where l represents the side length of the triangle.

So, we have:

$A_t = \frac{\sqrt 3}{4}x^2$

The area of the hexagonal base is 6 times the area of the triangle.

This is so because, an hexagon has 6 sides

So, we have:

$A_h = \frac{\sqrt 3}{4}x^2 \times 6$

$A_h = \frac{3\sqrt 3}{2}x^2$

Lastly, the volume of the pyramid is then calculated as:

$V = \frac 13 \times A_h \times h$

This gives

$V = \frac 13 \times \frac{3\sqrt 3}{2}x^2 \times 3x$

So, we have:

$V = \frac{3\sqrt 3}{2}x^2 \times x$

$V = \frac{3\sqrt 3}{2}x^3$

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